An isosceles triangle ABC is right angled at B.D is a point inside the triangle ABC. P and Q are the feet of the perpendiculars drawn from D on the sides AB and Ac respectively of $$\ \triangle ABC.\ $$If AP = a cm, AQ = b cm and $$\ \angle BAD$$= 15$$^\circ$$, sin $$\ 75^\circ$$=

$$\triangle$$ ABC is a right angled isosceles triangle right angled at B

Here $$\angle A = \angle C$$

$$90^\circ+2\angle A = 180^\circ$$

$$\therefore \angle A = \angle C = 45^\circ$$

Given $$\angle BAD = 15^\circ$$

From $$\triangle$$ ABC, $$\angle BAC = \angle BAD+\angle DAQ$$

$$\Rightarrow 45^\circ = 15^\circ+\angle DAQ$$

$$\therefore \angle DAQ = 30^\circ$$

From $$\triangle DAQ, \angle AQD = 90^\circ$$ and $$\angle DAQ = 30^\circ$$

$$\angle AQD+\angle DAQ+\angle ADQ = 180^\circ$$

$$90^\circ+30^\circ+\angle ADQ = 180^\circ$$

$$\Rightarrow \angle ADQ = 60^\circ$$

$$From \triangle ADQ$$, 

sin $$60^\circ = \frac{AQ}{AD}$$ 

$$\frac{\sqrt{3}}{2} = \frac{b}{AD}$$ ( $$\because$$ sin $$60^\circ = \frac{\sqrt{3}}{2})$$

AD = $$\frac{2b}{\sqrt{3}}$$

In $$\triangle APD, \angle APD = 90^\circ$$ and  $$\angle PAD = 15^\circ$$

$$\angle APD+\angle PAD+\angle ADP = 180^\circ$$

$$90^\circ+15^\circ+\angle ADP = 180^\circ$$

$$\Rightarrow \angle ADP = 75^\circ$$

From $$\triangle$$ APD, 

sin $$75^\circ = \frac{AP}{AD}$$

Substituting AD = $$\frac{2b}{\sqrt{3}}$$ in above equation

$$\Rightarrow$$ sin $$75^\circ = \frac{a}{(\frac{2b}{\sqrt{3}})}$$

$$\therefore$$  sin $$75^\circ = \frac{\sqrt{3}a}{2b}$$